Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(-\dfrac{3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)=\dfrac{-3}{17}+\dfrac{3}{17}+\dfrac{2}{3}=\dfrac{2}{3}\)
\(\dfrac{-5}{21}+\dfrac{-16}{21}+1=-1+1=0\)
\(\dfrac{-4}{11}\cdot\dfrac{5}{15}\cdot\dfrac{11}{-4}=\dfrac{5}{15}=\dfrac{1}{3}\)
A = \(\dfrac{2}{5}:\dfrac{3}{4}+\dfrac{2}{5}.\dfrac{3}{6}+\dfrac{2}{5}\)
A=\(\dfrac{2}{5}.\dfrac{4}{3}+\dfrac{2}{5}.\dfrac{3}{6}+\dfrac{2}{5}\)
A=\(\dfrac{2}{5}.\left(\dfrac{4}{3}+\dfrac{3}{6}+1\right)\) =\(\dfrac{2}{5}.\dfrac{17}{6}\)= \(\dfrac{17}{5}\)
B = \(\dfrac{5}{21}.\dfrac{4}{11}+\dfrac{7}{11}.\dfrac{5}{21}-\dfrac{2}{3}\)
B =\(\dfrac{5}{21}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)-\dfrac{2}{3}\)
B= \(\dfrac{5}{21}.1-\dfrac{2}{3}\) = \(\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{-3}{7}\)
a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)
\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}\)
\(=\frac{5}{66}\)
a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
ta có:
\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
...
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b,
\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)
ta có:
\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)
\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)
...
\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)
= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)
1/4.12/13+1/4.1/13-3/25
1/4.(12/13+1/13)-3/25
1/4.1-3/25
1/4-3/25
1/8
\(\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-12\%=\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-\frac{3}{25}=\frac{1}{4}\cdot\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)
\(=\frac{1}{4}\cdot1-\frac{3}{25}=\frac{1}{4}-\frac{3}{25}=\frac{13}{100}\)
Nhớ bài đây sửa đi sửa lại cũng vì do cái số " % " :(((
a) \(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)
Trường hợp 1 : \(\frac{2}{5}\) : x = \(\frac{1}{4}\)
=> x = \(\frac{2}{5}:\frac{1}{4}=\frac{2}{5}\cdot4=\frac{8}{5}\)
Trường hợp 2 : \(\frac{2}{5}:x=-\frac{1}{4}\)
=> \(x=\frac{2}{5}:\left(-\frac{1}{4}\right)=\frac{2}{5}\cdot\left(-4\right)=-\frac{8}{5}\)
Vậy \(x=\pm\frac{8}{5}\)
b) \(\frac{x}{24}=-\frac{1}{3}-\frac{1}{8}=-\frac{11}{24}\)
=> x = -11
c) \(\frac{3}{x+3}=\frac{-7}{21}\)
=> \(\frac{3}{x+3}=\frac{-1}{3}\)
=> -1(x + 3) = 9
=> -x - 3 = 9
=> -x = 12
=> x = -12
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
`a)1 4/23 + ( 5/21-4/23)+16/21-1/2`
`=27/23+5/21-4/23+16/21-1/2`
`=(27/23-4/23)+(5/21+16/21)-1/2`
`=23/23+21/21-1/2`
`=1+1-1/2`
`=2-1/2`
`=4/2-1/2`
`=3/2`
___
`b)75%-(5/2+5/3)+(-1/2)^3`
`=3/4-5/2+5/3+(-1/8)`
`=(3/4-5/2-1/8)+5/3`
`=(6/8-20/8-1/8)+5/3`
`=-15/8+5/3`
`=-45/24+40/24`
`=-5/24`
___
`c)-3/4(-55/9).8/11`
`=-3/4.(-40/9)`
`=-10/3`
__
`d)-3/8 . 6/13 + 7/13 . (-3/8) + 1 3/8`
`= -3/8 . (6/13 + 7/13) + 11/8`
`= -3/8 . 13/13 + 11/8`
`= -3/8 .1 + 11/8`
`= -3/8 + 11/8`
`= 8/8`
`=1`
\(\dfrac{21}{4}\cdot\dfrac{-11}{4}+\dfrac{21}{4}\cdot\dfrac{1}{2}+\dfrac{21}{4}\cdot\dfrac{-1}{2}\)
\(=\dfrac{21}{4}\cdot\dfrac{-11}{4}\cdot\dfrac{21}{4}+\dfrac{2}{4}+\dfrac{21}{4}\cdot\dfrac{-2}{4}\)
\(=\dfrac{21}{4}\cdot\left(\dfrac{-11}{4}+\dfrac{2}{4}+\dfrac{-2}{4}\right)\)
\(=\dfrac{21}{4}\cdot\dfrac{-11}{4}\)
\(=\dfrac{-231}{16}\)